Chapter 13, Problem 49P

To determine

The mass of lead should be added to give the diver neutral buoyancy.

Answer to Problem 49P

  $3.9\text{kg}$

Explanation of Solution

Given:

Mass of a diver, $m=85$ kg

Density of a diver, $\rho =0.96$ kg/L

Formula used:

Let, volume of the diver be $V$

Density of the diver be ${\rho }_D$

Volume of the added lead be $V_{Pb}$

Mass of the lead be $m_{Pb}$

Let us apply $\sum F_y=0$ to the diver,

  $B-W_D-W_{Pb}=0$  $\left(1\right)$

Where, $B$ is buoyant force

  $\begin{array}{l}B={\rho }_w{V}_{D+Pb}g\\ B={\rho }_w{V}_{D}g+{\rho }_w{V}_{Pb}g\end{array}$

  $W_D$ is weight of a diver

  $\begin{array}{c}{W}_D=m_{D}g\\ ={\rho }_D{V}_{D}g\end{array}$

  $W_{Pb}$ is weight of the lead, $W_{Pb}=m_{Pb}g$

Substituting for $B$ , $W_D$ , and $W_{Pb}$ in equation $\left(1\right)$ ,

  ${\rho }_w{V}_{D}g+{\rho }_w{V}_{Pb}g-{\rho }_D{V}_{D}g-m_{Pb}g=0$

Or

  ${\rho }_w{V}_D+{\rho }_w{V}_{Pb}-{\rho }_D{V}_D-m_{Pb}=0$  $\left(2\right)$

Where, ${\rho }_w$ is density of water,

  $V_D$ is volume of a diver,

  $g$ is acceleration due to gravity,

  $V_{Pb}$ is volume of the lead,

  ${\rho }_D$ is density of a diver,

  $m_{Pb}$ is mass of a lead.

Mass $\left(m\right)$ , density $\left(\rho \right)$ , and volume $\left(V\right)$ of a given substance is related through the expression,

  $m=\rho V\Rightarrow \rho =\frac{m}{V}$

Applying this to the equation $\left(2\right)$ ,

  ${\rho }_w\frac{m_D}{{\rho }_D}+{\rho }_w\frac{m_{Pb}}{{\rho }_{Pb}}-{\rho }_D\frac{m_D}{{\rho }_D}-m_{Pb}=0$

  $\Rightarrow {\rho }_w\frac{m_D}{{\rho }_D}-m_D+{\rho }_w\frac{m_{Pb}}{{\rho }_{Pb}}-m_{Pb}=0$

  $\Rightarrow m_D\left(\frac{{\rho }_w}{{\rho }_D}-1\right)+m_{Pb}\left(\frac{{\rho }_w}{{\rho }_{Pb}}-1\right)=0$

  $\Rightarrow m_D\left(\frac{{\rho }_w-{\rho }_D}{{\rho }_D}\right)+m_{Pb}\left(\frac{{\rho }_w-{\rho }_{Pb}}{{\rho }_{Pb}}\right)=0$

  $\Rightarrow m_{Pb}\left(\frac{{\rho }_w-{\rho }_{Pb}}{{\rho }_{Pb}}\right)=-m_D\left(\frac{{\rho }_w-{\rho }_D}{{\rho }_D}\right)$

or

  $-m_{Pb}\left(\frac{{\rho }_{Pb}-{\rho }_w}{{\rho }_{Pb}}\right)=-m_D\left(\frac{{\rho }_w-{\rho }_D}{{\rho }_D}\right)$

or

  $m_{Pb}\left(\frac{{\rho }_{Pb}-{\rho }_w}{{\rho }_{Pb}}\right)=m_D\left(\frac{{\rho }_w-{\rho }_D}{{\rho }_D}\right)$

  $\Rightarrow m_{Pb}=\frac{m_D\left(\frac{{\rho }_w-{\rho }_D}{{\rho }_D}\right)}{\left(\frac{{\rho }_{Pb}-{\rho }_w}{{\rho }_{Pb}}\right)}=\frac{{\rho }_{Pb}\left({\rho }_w-{\rho }_D\right)m_D}{{\rho }_D\left({\rho }_{Pb}-{\rho }_w\right)}$  $\left(3\right)$

Calculation:

Density of lead, ${\rho }_{Pb}=11.3\times {10}^3{\text{kg/m}}^{\text{3}}$

Density of water, ${\rho }_w=1.0\times {10}^3{\text{kg/m}}^{\text{3}}$

Substituting the numerical values in equation $\left(3\right)$ ,

  $\begin{array}{c}{m}_{Pb}=\frac{\left(\text{11}{\text{.3×10}}^{\text{3}}{\text{kg/m}}^{\text{3}}\right)\left({\text{1×10}}^{\text{3}}{\text{kg/m}}^{\text{3}}\text{-0}{\text{.96×10}}^{\text{3}}{\text{kg/m}}^{\text{3}}\right)\text{85 kg}}{\left(\text{0}{\text{.96×10}}^{\text{3}}{\text{kg/m}}^{\text{3}}\right)\left(\text{11}{\text{.3×10}}^{\text{3}}{\text{kg/m}}^{\text{3}}{\text{-1×10}}^{\text{3}}{\text{kg/m}}^{\text{3}}\right)}\\ =3.9\text{kg}\end{array}$

Conclusion:

The mass of lead should be added to give the diver neutral buoyancy $3.9\text{kg}$ .